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Chapter 2: Problem 30

Use transformations to graph the quadratic function and find the vertex of the associated parabola. $$g(s)=-(s-2)^{2}+2$$

Short Answer

Expert verified
The vertex of the parabola is at (2, 2) and the graph is a parabola opening downwards.

Step by step solution

01

Translate the Function

The quadratic equation given is already in the form \(f(x) = a(x-h)^2 + k\). Here, 'a' is -1, 'h' is 2 and 'k' is 2. This tells us that the basic quadratic function \(f(x) = x^2\) has been horizontally shifted right by 2 units, vertically shifted up by 2 units and then vertically reflected because 'a' is negative.
02

Identify the Vertex

The vertex of the parabola is given by the (h, k) values of -1 and 2 respectively which in this case is (2, 2).
03

Graph the Function

Begin by plotting the vertex (2, 2) on the graph. Since the quadratic is negatively oriented (opens down), draw the sides of the parabola so they open downwards from the vertex point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabola Transformations
Quadratic functions often come in the form of transformations. Understanding these transformations helps immensely in graphing them accurately. The function we have here is written as \(g(s)=-(s-2)^{2}+2\), which represents a lot of information about its graph.First, consider the form \(f(x) = a(x-h)^2 + k\). Here:
  • \(a\) represents vertical stretching or compressing, and when negative, it indicates a reflection over the x-axis.
  • \(h\) signifies the horizontal shift from the origin.
  • \(k\) indicates the vertical shift.
For \(g(s)=-(s-2)^{2}+2\):
  • The negative \(-1\) indicates the parabola is flipped upside down.
  • The \(s-2\) shows the graph shifts right by 2 units.
  • The \(+2\) indicates a vertical shift upwards by 2 units.
Transformations like these allow you to manipulate the basic shape of \(y=x^2\) to fit various applications.
Graphing Parabolas
Graphing a parabola involves a few key steps that help make the process simpler. Start by identifying essential transformations, then move on to plotting significant points like the vertex. Here's how you can graph \(g(s)=-(s-2)^2+2\):
  • Begin with the vertex, \( (2, 2) \), found from the transformation formula.
  • Next, consider the direction: since \(a=-1\) is negative, the parabola opens downwards.
  • Given the vertex as the starting point, plot symmetric points on either side. This helps visualize the curve efficiently.
Make sure to sketch the curve smoothly and ensure it conforms to the typical "U" shape known for parabolas, but inverted due to the negative sign. If done correctly, you'll have an accurate representation of the function on your graph.
Vertex of a Parabola
The vertex of a parabola is a crucial point that serves as either a peak or a dip on the graph. It is defined by the coordinates \( (h, k) \) in the function \(f(x) = a(x-h)^2 + k\). For our function, \(g(s)=-(s-2)^2+2\), the vertex is clearly given by \( (2, 2) \).The position of the vertex implies:
  • The "h" value (2) is where the parabola is horizontally located on the s-axis.
  • The "k" value (2) indicates the vertical placement on the same plane.
Graphically, this point stands as the highest point in a downward-opening parabola. Importantly, this vertex is not only a positional marker but also impacts the entire graph's shape and direction. When you have the vertex, you possess a key piece in accurately sketching the parabola.

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Most popular questions from this chapter

When designing buildings, engineers must pay careful attention to how different factors affect the load a structure can bear. The following table gives the load in terms of the weight of concrete that can be borne when threaded rod anchors of various diameters are used to form joints. $$\begin{array}{cc} \text { Diameter (in.) } & \text { Load (1b) } \\ \hline 0.3750 & 2105 \\ 0.5000 & 3750 \\ 0.6250 & 5875 \\ 0.7500 & 8460 \\ 0.8750 & 11,500 \end{array}$$ (a) Examine the table and explain why the relationship between the diameter and the load is not linear. (b) The function $$f(x)=14,926 x^{2}+148 x-51$$ gives the load (in pounds of concrete) that can be borne when rod anchors of diameter \(x\) (in inches) are employed. Use this function to determine the load for an anchor with a diameter of 0.8 inch. (c) since the rods are drilled into the concrete, the manufacturer's specifications sheet gives the load in terms of the diameter of the drill bit. This diameter is always 0.125 inch larger than the diameter of the anchor. Write the function in part (b) in terms of the diameter of the drill bit. The loads for the drill bits will be the same as the loads for the corresponding anchors. (Hint: Examine the table of values and see if you can present the table in terms of the diameter of the drill bit.)

Let \(f(x)=2 x+5\) and \(g(x)=f(x+2)-4 .\) Graph both functions on the same set of coordinate axes. Describe the transformation from \(f(x)\) to \(g(x) .\) What do you observe?

Sketch a graph of the quadratic function, indicating the vertex, the axis of symmetry, and any \(x\)-intercepts. $$F(s)=-2 s^{2}+3 s+1$$

In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=4(2 x+9)^{5}-(2 x+9)^{8}$$

Sketch a graph of the quadratic function, indicating the vertex, the axis of symmetry, and any \(x\)-intercepts. $$f(t)=-t^{2}-1$$
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